2
$\begingroup$

I am told:

$$\alpha + \beta - \gamma = \pi$$

And I have to prove:

$$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos \gamma$$

What should I be looking for? I kept trying to take the sine of bots sides and use the formulas:

$$\sin(a + b) = \sin a \cos b + \sin b \cos a$$

$$\sin(a-b) = \sin a \cos b - \sin b \cos a$$

but got nowhere. Then I tried using the formulas:

$$\sin a + \sin b = 2 \sin \bigg ( \dfrac{a + b}{2} \bigg ) \cos\bigg ( \dfrac{a - b}{2} \bigg )$$

$$\sin a - \sin b = 2 \cos \bigg ( \dfrac{a + b}{2} \bigg ) \sin \bigg ( \dfrac{a - b}{2} \bigg )$$

But again, I got nowhere. Can you give me a hint? At least what should I be looking for? What should be my strategy? Everything that I did felt just random, while kind of hoping that everything would just magically turn into the desired result. What is the strategy for this kind of problem?

$\endgroup$

3 Answers 3

2
$\begingroup$

Hint:

Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

$$\sin^2\beta-\sin^2\gamma=\sin(\beta+\gamma)\sin(\beta-\gamma)=\sin(\beta+\gamma)\sin(\pi-\alpha)=\sin(\beta+\gamma)\sin\alpha$$

Again,$$\sin^2\alpha=\sin\alpha\cdot\sin(\beta-\gamma)$$

Hope you can take it home from here?

$\endgroup$
1
$\begingroup$

$$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma-2 \sin \alpha \sin \beta \cos \gamma=$$

$$=\sin^2 \alpha + \sin^2 \beta -1+\cos^2(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)=$$ $$=\sin^2 \alpha-\cos^2 \beta+\cos(\alpha+\beta)\cos(\alpha-\beta)=$$ $$=\sin^2 \alpha-\cos^2 \beta+\cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta=$$ $$=(1-\sin^2\beta)\sin^2\alpha-(1-\cos^2\alpha)\cos^2\beta=0.$$

$\endgroup$
0
$\begingroup$

I can say that: $\sin(\alpha+\beta)=\sin(\pi +\gamma)$. Now, I can evaluate the expression: $$\sin(\alpha)^2+\sin(\beta)^2-\sin(\gamma)^2=\sin(\alpha)^2+\sin(\beta)^2-\sin(\alpha)^2\cos(\beta)^2-\sin(\beta)^2\cos(\alpha)^2-\sin(\alpha)\cos(\alpha)\sin(\beta)\cos(\beta)$$

Now substitute $\cos(\alpha)^2$ and $\cos(\beta)^2$, and obtain: $$2\sin(\alpha)\sin(\beta)(\sin(\alpha)\sin(\beta)-cos(\alpha)\cos(\beta))=2\sin(\alpha)\sin(\beta)(-1\cdot\cos(\alpha+\beta))=2\sin(\alpha)\sin(\beta)\cos(\gamma)$$

Also: $$\cos(\gamma)=\cos(\alpha+\beta-\pi)=\cos(\pi-(\alpha+\beta))=-\cos(\alpha+\beta)$$

From here, as you wanted: $$\sin(\alpha)^2+\sin(\beta)^2-\sin(\gamma)^2=2\sin(\alpha)\sin(\beta)\cos(\gamma)$$

$\endgroup$
2
  • $\begingroup$ But isn't it that $\cos(a + b) = \cos a \cos b - \sin a \sin b$? So shouldn't we have in your second to last line $\sin(\alpha) \sin(\beta) - \cos(\alpha) \cos(\beta) = - \cos(\alpha + \beta)$ instead of $\cos(\alpha + \beta)$? $\endgroup$
    – user592938
    Jan 27, 2020 at 22:10
  • $\begingroup$ @user1502: see my edits. $\endgroup$
    – Matteo
    Jan 28, 2020 at 19:38

You must log in to answer this question.